4x+3y=x^2-xy=8

Question

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Answer 1

4x+3y=x²-xy, so 3y+xy=x²-4x, y(x+3)=x²-4x, y=(x²-4x)/(x+3).

4x+3y=8, so, substituting for y, 4x+3(x²-4x)/(x+3)=8.

Multiply through by x+3:

4x²+12x+3x²-12x=8x+24,

7x²-8x-24=0.

Using the quadratic formula: x=(8±√(64+672))/14=(8±√736)/14=(4+2√46)/7 or (4-2√46)/7.

So y=(8-4x)/3=(40-8√46)/21 or (40+8√46)/21.

So there are two (x,y) solutions: ((4+2√46)/7,(40-8√46)/21) or ((4-2√46)/7,(40+8√46)/21).

That is, (2.5092,-0.6790) or (-1.3664,4.4885) approx.